On the set $K_3(G)$ of finite groups elements commuting exactly with three elements

Let $G$ be an arbitrary finite multiplicative group, $|G| = n$. We define the set $K_3(G)$ as follows:
$$ K_3(G) = \{x \in G \mid |C_G(x)| = 3\} = \{x \in G \mid C_G(x) = \{e, x, x^2\}\}. $$
It follows from the definition of $K_3(G)$ that
A) if $x \in K_3(G)$, then the order of $x$ is $3$ ($o(x) = 3$);
B) if $x \in K_3(G)$, then $x^2 \in K_3(G)$.
The following properties of the set $K_3(G)$ have been proved.
Proposition 1. If $K_3(G) \ne\varnothing$, then $|G| \,\vdots\, 3$ and $|G| \not\,\vdots\, 9$.
Proposition 2. If $x \in K_3(G)$, then $x^g \in K_3(G)$ for each $g \in G$.
Proposition 3. Let $K_3(G) \ne\varnothing$, $x \in G$ and $o(x) = 3$. Then $x \in K_3(G)$.
Proposition 4. Let $|G| = n; K_3(G) \ne\varnothing$. Then $|K_3(G)| \in \left\{\frac n3;\frac{2n}3\right\}$.
Lemma 5. Let $a, g \in G$, $o(a) = 3$; $g^{-1}ag = a^2$. Then $o(g)\,\vdots\, 2$.
Proposition 6. 1) Let $o(a) = 3$ and $g^{-1}ag = a^2$. Then $|G|\,\vdots\, 6$.
2) If $|G| = 2k + 1$, then $K_3(G) = \varnothing$ or $|K_3 (G)| =\frac{2|G|}3$.
Theorem 7. Let $G$ be a finite simple group, $|G| = n$, $K_3(G) \ne\varnothing$. Then all involutions of the group $G$ form a class of conjugate elements.